Find the shaded area

The area of $\triangle ACP$ is given by:

$\begin{aligned} [ACP] & = [ABC]-[BCP]-[ABP] \\ & = \frac {20\times 10}2 - \frac {6\times 8}2 - \frac 12 \times AB \times BP \sin \blue{\angle ABP} & \small \blue{\text{Note that }\angle ABP = \angle BCP} \\ & = 100 - 24 - \frac {20 \times 6 \times \blue{0.6}}2 & \small \blue{\text{and }\sin \angle BCP = \frac 6{10}} \\ & = 100 - 24 - 36 = \boxed{40} \end{aligned}$

Let $\angle PBC=\theta$ and $\angle PCB=\beta$ , then

$\angle PBA=\angle ABC-\angle PBC=90^\circ - \theta = \beta$

So $\angle PAB=\theta$ and $\triangle APB$ is a right triangle.

$\cos \theta=\dfrac{AP}{20}=\dfrac{6}{10}$

$AP=12$

The area of the shaded region colored blue is $\dfrac{1}{2}[20\times10-8\times 6-6\times 12]=\boxed{40}$

$sin( \angle{PBA})=\frac{6}{10}$

Then the area is simply ,

$\frac{1}{2} *20*10 - \frac{1}{2} *8 * 10 - \frac{1}{2} *6*20* \frac{6}{10} = 40$

$|\overline {AC}|=\sqrt {20^2+10^2}=10\sqrt 5$ .

$\tan \angle {ACB}=2,\tan \angle {PCB}=\dfrac{3}{4}\implies \tan \angle {ACP}=\dfrac{2-\frac{3}{4}}{1+\frac{3}{2}}=\dfrac{1}{2}\implies \sin \angle {ACP}=\dfrac{1}{\sqrt 5}$ .

So, area of $\triangle {ACP}=\dfrac{1}{2}\times |\overline {AC}|\times |\overline {PC}|\times \sin \angle {ACP}=\dfrac{1}{2}\times 10\sqrt 5\times 8\times \dfrac{1}{\sqrt 5}=\boxed {40}$ .